Rayleigh flow: flow with heat addition

Why it matters

Combustion chambers, afterburners, ramjet and scramjet combustors all add heat to a moving stream in a duct of roughly constant area. Rayleigh flow shows how much heat the stream can accept before it chokes, why heat addition always costs stagnation pressure, and why supersonic combustion behaves so differently from subsonic combustion.

Key ideas

Model. Steady, one-dimensional flow of a perfect gas in a constant-area duct, frictionless, with heat q added (or removed) per unit mass and no work. Heat release by combustion is modelled as external heating; changes in gas composition and γ are ignored.

Governing equations.

• Mass: ρ·V = constant.
• Momentum (no friction, constant area): p + ρ·V² = constant, i.e. p·(1 + γM²) = constant.
• Energy: q = cp·(T₀₂ − T₀₁). Heat addition raises T₀; that is its only direct effect on the energy equation.

Rayleigh line. Plotting the states allowed by mass and momentum on the T–s diagram gives the Rayleigh line. Two key points:

• Maximum entropy at M = 1. Adding heat (which raises entropy) drives the flow towards M = 1 from either side; cooling drives it away.
• Maximum static temperature at M = 1/√γ (0.845 for air). Between M = 0.845 and M = 1, adding heat actually lowers the static temperature, because the kinetic-energy rise exceeds the heat added.

Trends with heat addition.

Quantity	Subsonic (M < 1)	Supersonic (M > 1)
M	increases	decreases
p	decreases	increases
V	increases	decreases
ρ	decreases	increases
T₀	increases	increases
p₀	decreases	decreases
T	increases (decreases for 0.845 < M < 1)	increases

Thermal choking. The maximum heat that can be added for given inlet conditions is the amount that brings the exit to M = 1: q_max = cp·(T₀* − T₀₁). If more heat is added:

• subsonic inlet: the flow adjusts upstream — the inlet Mach number and mass flow drop;
• supersonic inlet: a normal shock forms upstream in the duct (or the intake unstarts), making the flow subsonic.

Stagnation-pressure loss. Heating always lowers p₀. At low Mach numbers the loss is small (Δp₀/p₀ ≈ (γ/2)M²·ΔT₀/T₀), which is why combustors are designed with low entry Mach numbers (about 0.2–0.3). Supersonic heat addition costs much more p₀.

Comparison with Fanno flow. Both use constant area and tend to M = 1, but Fanno keeps T₀ constant with friction, while Rayleigh changes T₀ without friction.

Formulas

p/p* = (1 + γ) / (1 + γ·M²)

T/T* = M²·(1 + γ)² / (1 + γ·M²)²

ρ*/ρ = V/V* = (1 + γ)·M² / (1 + γ·M²)

T₀/T₀* = (γ + 1)·M²·(2 + (γ − 1)·M²) / (1 + γ·M²)²

p₀/p₀* = ((1 + γ)/(1 + γ·M²))·[ (2 + (γ − 1)·M²)/(γ + 1) ]^(γ/(γ−1))

q = cp·(T₀₂ − T₀₁) (J/kg); between two sections use ratios of starred values, e.g. p₂/p₁ = (p/p*)₂/(p/p*)₁.

s₂ − s₁ = cp·ln(T₂/T₁) − R·ln(p₂/p₁) (J/(kg·K)).

Symbols: superscript * = state at M = 1 on the same Rayleigh line (same mass flux and impulse); q heat added per unit mass (J/kg); cp (1004.5 J/(kg·K) for air); T₀, p₀ stagnation temperature (K) and pressure (Pa); M Mach number; γ ratio of specific heats.

Worked examples

Example 1 (standard). Air enters a constant-area combustor at M₁ = 0.25, T₁ = 500 K, p₁ = 300 kPa. Find the heat addition that chokes the duct, and the exit p and T. γ = 1.4, cp = 1004.5 J/(kg·K).

1. T₀₁ = 500 × (1 + 0.2 × 0.0625) = 506.25 K.
2. T₀₁/T₀* = 2.4 × 0.0625 × (2 + 0.4 × 0.0625)/(1 + 1.4 × 0.0625)² = 0.30375/1.18266 = 0.2568.
3. T₀* = 506.25/0.2568 = 1971 K.
4. q_max = 1004.5 × (1971 − 506.25) = 1.47 × 10⁶ J/kg.
5. p₁/p* = 2.4/1.0875 = 2.207, so p* = 300/2.207 = 135.9 kPa. T₁/T* = 0.0625 × 5.76/1.18266 = 0.3044, so T* = 500/0.3044 = 1643 K.

Answer: q_max ≈ 1.47 MJ/kg; exit p ≈ 136 kPa, T ≈ 1643 K.

Example 2 (GATE level). Air at M₁ = 0.3, T₀₁ = 400 K receives q = 500 kJ/kg in a constant-area duct. Find M₂ and p₀₂/p₀₁.

1. T₀₂ = 400 + 500 000/1004.5 = 897.8 K.
2. (T₀/T₀*)₁ at M = 0.3 = 0.3469, so (T₀/T₀*)₂ = 0.3469 × 897.8/400 = 0.7785.
3. Inverting (Rayleigh table, subsonic branch): M₂ = 0.565.
4. (p₀/p₀*)₁ = 1.1985, (p₀/p₀*)₂ = 1.0882 at M = 0.565.
5. p₀₂/p₀₁ = 1.0882/1.1985 = 0.908.

Answer: M₂ ≈ 0.57, p₀₂/p₀₁ ≈ 0.91 — a 9% stagnation-pressure loss from heating alone.

Common mistakes

• Assuming T₀ is constant (that is Fanno flow) or using isentropic relations between inlet and exit.
• Expecting heat addition to always raise static temperature; between M = 0.845 and 1 it lowers T.
• Thinking heating can take subsonic flow past M = 1 in a constant-area duct; it chokes at M = 1.
• Dividing starred ratios the wrong way round when going between two sections.
• Using Δs = cp ln(T₂/T₁) without the pressure term.
• Forgetting that q is per unit mass; multiply by ṁ for heat rate (W).

For GATE AE

Expect: exit Mach number for a given heat addition, maximum heat before choking, stagnation-pressure loss, ratios of p, T, V between sections from tables, the M = 1/√γ temperature peak, and true/false trend questions for subsonic versus supersonic heating. Practise using Rayleigh tables with the ratio-of-ratios method.

Quick check

1. What does heat addition do to M in a supersonic Rayleigh flow?
2. At what Mach number is static temperature maximum on the Rayleigh line for air?
3. Does p₀ rise or fall with heating?
4. What is p/p* at M = 0 for air?
5. What happens if more than q_max is added to a subsonic flow?

Answers: 1. it decreases towards 1; 2. 0.845; 3. falls; 4. 2.4; 5. the inlet Mach number and mass flow decrease (thermal choking).